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21 Jun 2026, 10:31
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
60% (02:58) correct
40%
(02:50)
wrong
based on 78
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A pool has 15 inlet pipes. Each pipe can be set to low or high or can be left off. A pipe on low fills the pool in 12 hours, while a pipe on high fills the pool in 5 hours. Yesterday, the pool was filled from empty in exactly 40 minutes. How many pipes were left off?
A. 2
B. 4
C. 6
D. 7
E. 8
A. 2
B. 4
C. 6
D. 7
E. 8
21 Jun 2026, 21:23
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Here's my solution:
Note: x & y are number of high & low pipes respectively.
Note: x & y are number of high & low pipes respectively.
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IMG_20260622_095239.jpg [ 276.21 KiB | Viewed 510 times ]
22 Jun 2026, 05:52
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This is good mathematical deduction.
I am diving straight into how I solved it.
First 1/5 = 0.2.
1/12 gives recurring decimal. Is that helpful? NO. Why?
Because:
2/3(x) = 1.
x = 3/2.
We need a combination a 1/12 and 1/5 where we land 1.5
Considering any multiple of 1/5 gives only 1 decimal point, we need a multiple of 1/12 as well that gives us only 1 decimal point ideally.
If we quickly scan its multiples we see that 6/12 = 0.5 is the only such multiple.
Thus We have 6 of the low filling pipes and the remaining which is 5 of high filling pipes. How?
1.5-0.5 = 1 [Remaining of the 3/2 we calculated]
1/5*5 = 1.
Thus 5 high filling pipes.
6+5 = 11 and 4 are left off.
I am diving straight into how I solved it.
First 1/5 = 0.2.
1/12 gives recurring decimal. Is that helpful? NO. Why?
Because:
2/3(x) = 1.
x = 3/2.
We need a combination a 1/12 and 1/5 where we land 1.5
Considering any multiple of 1/5 gives only 1 decimal point, we need a multiple of 1/12 as well that gives us only 1 decimal point ideally.
If we quickly scan its multiples we see that 6/12 = 0.5 is the only such multiple.
Thus We have 6 of the low filling pipes and the remaining which is 5 of high filling pipes. How?
1.5-0.5 = 1 [Remaining of the 3/2 we calculated]
1/5*5 = 1.
Thus 5 high filling pipes.
6+5 = 11 and 4 are left off.
kevincan
23 Jun 2026, 00:30
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Let's assume tank to be of 60 L (LCM of 12 and 5)
Fill rate at low flow = 60/12 = 5L/h
Fill rate at High Flow = 60/5 = 12L/h
Time to fill = 40 min= 2/3 h
Let there be X low flow pipe opened and Y High flow pipes opened.
=>
(5X + 12Y) x (2/3) = 60 L
5X + 12 Y = 90
The only possible solution for X and Y if they are positive and integer:
x= 6, y = 5
Total open piper = 11
Fill rate at low flow = 60/12 = 5L/h
Fill rate at High Flow = 60/5 = 12L/h
Time to fill = 40 min= 2/3 h
Let there be X low flow pipe opened and Y High flow pipes opened.
=>
(5X + 12Y) x (2/3) = 60 L
5X + 12 Y = 90
The only possible solution for X and Y if they are positive and integer:
x= 6, y = 5
Total open piper = 11
